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Math Help - Determinant, Change of basis

  1. #1
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    Determinant, Change of basis

    Hi!

    Let V be a vector space, \dim (V)=n and \omega \in \mathrm{Alt}^n V,\ \omega \neq 0 an alternating form. Let M be the change of basis matrix from the basis a=(a_1,...,a_n) for V to the basis  b=(b_1,...,b_n) for V.

    Proof that \det M=\frac{\omega(a_1,...,a_n)}{\omega(b_1,...,b_n)}<br />

    I would like to use the Leibniz formula
    \det M = \sum_{\sigma \in S_n} \mathrm{sign} (\sigma) \cdot M_{1 \sigma(1)}\cdot ... \cdot M_{n \sigma(n)}

    Since a=(a_1,...,a_n) and b=(b_1,...,b_n) are bases for V, I can write

     a_i = \sum_{j=1}^{n}M_{ij}b_j
    and therefore

    \frac{\omega(\sum_{j=1}^{n}M_{1j}b_j,...,  \sum_{j=1}^{n}M_{nj}b_j)}{\omega(b_1,...,b_n)}


    =\frac{\omega( M_{11}b_1+...+M_{1n}b_n ,..., M_{n1}b_1+...+M_{nn}b_n )}{\omega(b_1,...,b_n)}

    How should I continue? Can I write the last as \frac{\sum_{j=1}^n M_{1j} \cdot ... \cdot M_{nj} \cdot \omega(b_1,...,b_n)}{\omega(b_1,...,b_n)} because \omega is alternating?

    Thanks in advance!

    Bye,
    Lisa
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  2. #2
    MHF Contributor

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    Quote Originally Posted by lisa View Post
    Hi!

    Let V be a vector space, \dim (V)=n and \omega \in \mathrm{Alt}^n V,\ \omega \neq 0 an alternating form. Let M be the change of basis matrix from the basis a=(a_1,...,a_n) for V to the basis  b=(b_1,...,b_n) for V.

    Proof that \det M=\frac{\omega(a_1,...,a_n)}{\omega(b_1,...,b_n)}<br />

    I would like to use the Leibniz formula
    \det M = \sum_{\sigma \in S_n} \mathrm{sign} (\sigma) \cdot M_{1 \sigma(1)}\cdot ... \cdot M_{n \sigma(n)}

    Since a=(a_1,...,a_n) and b=(b_1,...,b_n) are bases for V, I can write

     a_i = \sum_{j=1}^{n}M_{ij}b_j
    and therefore

    \frac{\omega(\sum_{j=1}^{n}M_{1j}b_j,..., \sum_{j=1}^{n}M_{nj}b_j)}{\omega(b_1,...,b_n)}


    =\frac{\omega( M_{11}b_1+...+M_{1n}b_n ,..., M_{n1}b_1+...+M_{nn}b_n )}{\omega(b_1,...,b_n)}

    How should I continue? Can I write the last as \frac{\sum_{j=1}^n M_{1j} \cdot ... \cdot M_{nj} \cdot \omega(b_1,...,b_n)}{\omega(b_1,...,b_n)} because \omega is alternating?

    Thanks in advance!

    Bye,
    Lisa
    the point here is that since \omega is alternating we have \omega(x_1, \cdots , x_n)=0 if x_i=x_j for some i \neq j. so using multilinearity of \omega to expand \omega(\sum_{j=1}^{n}M_{1j}b_j,..., \sum_{j=1}^{n}M_{nj}b_j), all terms in which one of b_j appears more than once will be zero. thus we'll be left with terms in the form c_{\sigma} \omega(b_{\sigma(1)}, \cdots , b_{\sigma(n)}), where \sigma \in S_n and c_{\sigma} is in terms of M_{ij}. again, using the fact that \omega is alternating, we have

     \omega(b_{\sigma(1)}, \cdots , b_{\sigma(n)})=\text{sgn}(\sigma) \cdot \omega(b_1, \cdots , b_n). the only thing you need to show now is that c_{\sigma}=M_{1\sigma(1)} \cdot \cdots M_{n \sigma(n)}. (left for you!)
    Last edited by NonCommAlg; May 17th 2010 at 05:03 AM.
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